# How can Navier--Stokes equations have asymmetric solutions such as Karman vortex streets

The Navier--Stokes equations are axially symmetric, so with symmetric boundary conditions, how can features such as Karman vortex streets develop?

I understand that in reality symmetry does never completely hold due to minimal fluctuations of geometries, densities, etc., but in the linked article, vortex streets develop even in numerical simulations

• No. in this case you have no symmetry boundary condition on the center of the cylinder (well circle). The circle is fully meshed. The instability develops because of the non-linearity of the Navier-Stokes equation. For a given Reynolds number, the solution without oscillations around the cylinder is unstable from a stability analysis point of view. Consequently, slight perturbations in the numerical solution (round-off errors, meshing, order of operations) start-off the oscillations. – BlaB Aug 9 '18 at 14:11
• A good example of stability analysis of a flow is the Rayleigh-Bénard instability. You can demonstrate via linear stability analysis that for a given critical Rayleigh number the absence of flow is unstable to any slight perturbation $\epsilon$. As such any type of rounding error in your numerical scheme is enough to generate this $\epsilon$ and start-off your Rayleigh-Bénard instability. – BlaB Aug 9 '18 at 14:12
• I am not sure what you mean by a symmetric numerical method. Somewhere somehow you will always do operations in a certain order, whether it is when solving a numerical system by an iterative (or direct) solver or by looping over certain cell in a certain order. This leads to some roundoff errors which are on the order of machine precision (say 1e-16). You could have a symmetric boundary condition at the center of the cylinder by meshing half the cylinder and imposing slip on that surface. In this way you would constraint your space to half a cylinder and the instability would not appear. – BlaB Aug 9 '18 at 14:20
• The Navier-Stokes equation (or at least the pressure part of them) generally are solved implicitely. If you had a perfectly symmetric numerical method where the order of the operations and all your parenthesis were balanced, then you would not get this instability without manually introducing a slight perturbation in your solution. In real life, you always have a source of floating point error which appears due to a number of reason and then propagates. Furthermore, I fail to see how you could perfectly mesh a circle with an ideally symmetric finite difference method... – BlaB Aug 9 '18 at 14:29
• That's all I wanted to reconfirm, thanks. (BTW, I don't understand the last bit. I never talked about PERFECTLY meshing anything, just about symmetrically meshing things) – Bananach Aug 9 '18 at 17:47